Do you mean an adjustable current limit?Just as a side question, how would I make the output amperage variable?
I think you need a bigger filter cap to be happy. Below shows the ripple at 1.5 amp load with 1000ufd. and 6800 ufd.. You need to allow 2.5 volts for the regulator so 6800 ufd. would let you get to 12 volts on the low tap at 1.5 amps. Now for the second part: Don't worry to much about the current limit, the IC has built in thermal protection, so if it gets to hot it will shut down. Look up the data sheet there will be some info there. So how much current can you expect? Well it depends on if you use a heatsink. The part in free air can dissipate about 2.5 watts. Now mind you this puts the junction at 150C. (50 degrees C per watt) So with no heatsink you have about 15 volts into the regulator and 12 volts out so it can supply .8 amp maximum. .8 amp X 3 volts = 2.4 watts. Using the same math at 6 volts out you can supply .2 amps at 25C ambient. Not enough I bet. So add a heatsink, lets ay 6 degree C per watt. Now you have to look at thermal resistance junction to case. For the TO-220 part this is 5C per watt. Add this to the 6C for the heatsink and you have 11C per watt.
Now your cooking! You can dissipate 11 watts. Now you can get full capacity at 12 volts and about 1.2 amps at 6 volts. A better heatsink or a fan and you can get more at the lower voltages. The same ratios will apply with the high side switch. Full output at about 24 volts and less as you go down to 12.
https://www.electro-tech-online.com/custompdfs/2010/08/LM117-2.pdf
Hope this helps.
An LM317 cannot supply 1.5A when it has an input to output voltage of 13V or more. It gets hotter than a huge heatsink can cool it. A high velocity fan or liquid nitrogen will help.
The calculation of the resistor values and their power dissipation requires grade 4 arithmatic.
There is an easier way to get half DC voltage from the C.T. transformer and a bridge rectifier.
Leave the bridge rectifier hooked up without the switches you have. The full DC voltage will be present on the positive output of bridge rectifier. Half the DC voltage will be present at the center tap of the transformer. You can put filter caps on both points for two outputs.
In the C.T. DC output configuration, the two rectifiers to ground in the bridge on transformer outside terminal connections make the full wave rectifier for the half voltage output.
When output of regulator is set for lower voltage you can SPDT switch to the transformer center tap to reduce the voltage drop across the regulator and therefore its power dissipation.
As mentioned, you need a lot larger filter cap then 1000 uF for amperage you are trying for.
... I just don't know how to limit the current... would I put an appropriate resistor on the + output?
Look at the many heatsinks at a good electronic parts distributor like Digikey or Newark. They have links to manufacturers' sites who show sizes.How big would a heat sink with a thermal resistance of 6°C/W be?
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